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3 years ago

Three resistors of 3.3 kΩ, 4.7 kΩ and 3.9 kΩ are connected in series across a

Physics

1 answer:

natima [27]3 years ago

6 0

Explanation:

A.) equivalent resistance

Equivalent resistance ( Rt ) in a series connection is sum of all resistance.

Rt = 3300 + 4700 + 3900

= 11900 ohms

= 11.9 k ohm

B.) current is constant across the whole circuit if it is in series .

current = potential difference ÷ total resistance

current = 12 / 11900 = 0.00100 Amphere

= 0.1 mA

C. potential difference ( v ) across the circuit =

current( i ) × resistance ( Rt )

=》v = 0.1 × 0.001 × 11900 = 12 volts

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A roller coaster moves 200 ft horizontally, then rises 135 ft at an angle of 30.0Â° above the horizontal. Next, it travels 135 f

vekshin1

Answer:

171 ft

Explanation:

The distance to be calculated is AB

AB = AC + BC

AC; Cos 60 = AC ÷ 135

AC = 135 cos 60 = 67.5 ft

BC; Cos 40 = BC ÷ 135

BC = 135 cos 40 = 103.416 ft

AB = 67.5 + 103.416 = 170.9159998210 ft

Distance between starting point to end point = 171 ft

4 0

3 years ago

PLSS HELP NOWW!!!!!!(01.01 LC) Which is an example of the force of attraction between two objects that have mass? (2 points) a M

Lyrx [107]

Answer:

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6 0

3 years ago

In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho

Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

$R=u_xt=(ucos \theta)t$

Therefore,

$t=\frac{R}{ucos\theta} .......(1)$

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

$y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2$

Substitute the value of t from equation (1).

$y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )$

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

$y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s$

The shot put was thrown with a speed 15.02 m/s.

7 0

3 years ago

Give proof that mass is constant and weight keeps changing.

shutvik [7]

The mass of an object stays the same wherever it is, but its weight can change. This happens if the object goes where the gravitational field strength is different from the gravitational field strength on Earth, such as into space or another planet.

7 0

2 years ago

Balance the following chemical equations

IrinaVladis [17]

Answer:

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Explanation:

4 0

3 years ago